1. ## Law of Vectors

The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below.

| a × b |² + (a • b)² = |a|²|b|²

My work, LSH:

= | a × b |² + (a • b)²

= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)

= (|a|²)(|a||b|)(|a|sinθ)(|a||b|)(|b|²)(|b|sinθ)(|a| sinθ)(|b|sinθ)(sin²θ) + (|a|²)(|a||b|)(|a|cosθ)(|a||b|)(|b|²)(|b|cosθ)(|a| cosθ)(|b|cosθ)(cos²θ)

= (|a|²)(|a||b|)²(|a|sinθ)²(|b|²)(|b|sinθ)²(sin²θ) + (|a|²)(|a||b|)²(|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)

= (|a|²|b|²(|a||b|)²) [(|a|sinθ)²(|b|sinθ)²(sin²θ) + (|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)]

= (|a|²|b|²(|a||b|)²) [(|a|²)(sin²θ)(|b|²)(sin²θ)(sin²θ) + (|a|²)(cos²θ)(|b|²)(|b|²)(cos²θ)(cos²θ)]

= (|a|²|b|²(|a||b|)²) [(sin²θ)(sin²θ)(sin²θ) + (cos²θ)(cos²θ)(cos²θ)]

And now I don't know what else to do! Please help. Did I mess up somewhere in my steps? Or is it possible to common factor still?

2. Originally Posted by Macleef
The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below.

| a × b |² + (a • b)² = |a|²|b|²

My work, LSH:

= | a × b |² + (a • b)²

= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)
I'm not sure how you got your third line, but continuing from the second:
$= |a|^2 |b|^2 (sin^2(\theta) + cos^2(\theta))$

$= |a|^2 |b|^2$

-Dan

Hi there,

= | a × b |² + (a • b)²

= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)

it's good up to here but then you're problem was how you multiplied out the brackets, remember multiplication distributes over addition, so if you had
(|a|+|b|+sinθ)(|a|+|b|+sinθ) you would have been multiplying them out the correct way (ish)... however you just have a product, so just stick them together and carry on, the last step given by topsquark follows from the trigonometric identity cos^2 +sin^2=1 which can be proved using pythagoras theorem and the unit circle (ask me or search the web if you are interested..)

4. $
\left( {\vec a \times \vec b} \right)^2 + \left( {\vec a \bullet \vec b} \right)^2 = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2
$

$
\left( {\left\| {\vec a} \right\|\left\| {\vec b} \right\|\sin (\theta )\vec n} \right)^2 + \left( {\left\| {\vec a} \right\|\left\| {\vec b} \right\|\cos (\theta )} \right)^2 = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2
$

$
\left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2 \sin ^2 (\theta )\left( {\vec n \times \vec n} \right) + \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2 \cos ^2 (\theta ) = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2$

$\left( {\sin ^2 (\theta ) + \cos ^2 (\theta )} \right)\left( {\left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2 } \right)\left( {\left( {\vec n \times \vec n} \right) + 1} \right) = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2
$

$
\left( 1 \right)\left( {\left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2 } \right)\left( {\left( 0 \right) + 1} \right) = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2
$

$
\left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2 = \left\| {\vec a} \right\|^2 \left\| {\vec b} \right\|^2
$

5. Why not use known vector identities?
$\left( {a \times b} \right) \cdot \left( {c \times d} \right) = \left( {a \cdot c} \right)\left( {b \cdot d} \right) - \left( {a \cdot d} \right)\left( {b \cdot c} \right)$

Thus, $\left\| {\left( {a \times b} \right)} \right\|^2 = \left( {a \times b} \right) \cdot \left( {a \times b} \right) = \left( {a \cdot a} \right)\left( {b \cdot b} \right) - \left( {a \cdot b} \right)^2$.
That gives the result.